S.O.S. Mathematics CyberBoard

[KnuthFan19873]

Calculate the line integral where C is the circle (x-1)^2+y^2=1 (go counterclockwise).

Use pullback, not Green's Theorem.

I parametrized the region with with t from 0 to 1 and with t from 0 to 1. Is that the best way to do it?

[Marvin_M]

[KnuthFan19873]

But its not centered at zero?

[dmoran]

What is pullback? I've never heard the term.

Dave
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[KnuthFan19873]

Go here.

[Zathras]

This circle can be expressed as r = 2 cos(theta). Theta is your parameter.

[KnuthFan19873]

Then, how do you change a differential 1-form to polar?

[commutative]

x - 1 = cos(t), and y = sin(t), 0 <= t <= 2pi. then dx = - sin(t)dt and dy = cos(t)dt.
hence: xdy - ydx = (1 + cos(t))dt, and sqrt(x^2 + y^2) = 2|cos(t/2)|. so you just
need to find the integral of 2(1 + cos(t))|cos(t/2)|, t from 0 to 2pi. this equals the
integral of 4(1 + cos(t))cos(t/2), t from 0 to pi, which is very easy to evaluate. the
final answer is 32/3.

[KnuthFan19873]

Well done commutative. I also tried to do it with Green's Theorem and ran into the following problem:

I found that (omega is the original integrand). So, after converting to polar coordinates, Green's Theorem gives which equals zero unless you change the limits on the outer integral to -pi/2 to pi/2 in which case you get the right answer.

Why is this necessary?

Also, just curious, how can you just "get rid" of the wedge product of dx and dy when you do these kinds of problems?

[qleak]

[commutative]

theta is obviously between -pi/2 and pi/2 and not 0 and pi.
note that half of the circle is in the first and the other half in the fourth
quadrant. if O is the origin (not the center of the circle) and A a point
inside (or on) the circle, then r = |OA| and theta is the angle between
x-axis and OA. to understand why theta is between -pi/2 & pi/2 better,
just choose some points in the circle, first or fourth quadrant, and see
how theta changes.